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\(HB.HC=15^2=225\)
Ta có : \(\hept{\begin{cases}AB^2=BH.BC\\AC^2=CH.BH\end{cases}\Rightarrow\frac{AB^2}{AC^2}=\frac{BH}{CH}\Rightarrow\hept{\begin{cases}\frac{HB}{HC}=\frac{25}{49}\\HB.HC=225\end{cases}\Rightarrow}\hept{\begin{cases}HB.HC.\frac{HB}{HC}=\frac{25}{49}.225\\HB.HC=225\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}HB^2=\frac{5625}{49}\\HB.HC=225\end{cases}\Rightarrow\hept{\begin{cases}HB=\frac{75}{7}\\HC=21\end{cases}}}\)
Xét tam giác ABC vuông tại A có AH là đường cao:
+) \(\tan C=\dfrac{AB}{AC}\) (TSLG)
\(\Rightarrow\tan C=\dfrac{3}{4}\Rightarrow\widehat{C}\approx37^0\)
\(\Rightarrow\widehat{B}=90^0-\widehat{C}\approx90^0-37^0\approx53^0\)
+) \(\sin C=\dfrac{AB}{BC}\) (TSLG)
\(\Rightarrow\sin37^0=\dfrac{AB}{20}\Rightarrow AB\approx12\) (cm)
+) \(AB^2+AC^2=BC^2\) (ĐL Pytago)
\(\Rightarrow AC=\sqrt{BC^2-AB^2}\approx\sqrt{20^2-12^2}\approx16\) (cm)
+) \(AB^2=BH.BC\) (HTL)
\(\Rightarrow BH=\dfrac{AB^2}{BC}\approx\dfrac{12^2}{20}\approx7,2\) (cm)
+) \(BH+CH=BC\)
\(\Rightarrow CH=BC-BH\approx20-7,2\approx12,8\) (cm)
Vậy \(HB\approx7,2cm;HC\approx12,8cm\)
Ta có: \(\dfrac{HB}{HC}=\dfrac{9}{16}\Rightarrow HB=\dfrac{9}{16}HC\)
Ta có: \(AB^2=BH.BC=BH\left(BH+HC\right)=\dfrac{9}{16}HC\left(\dfrac{9}{16}HC+HC\right)\)
\(=\dfrac{9}{16}HC.\dfrac{25}{16}HC=\dfrac{225}{256}HC^2\)
\(\Rightarrow HC^2=\dfrac{256AB^2}{225}=\dfrac{16384}{25}\Rightarrow HC=\dfrac{128}{5}\left(cm\right)\)
\(\Rightarrow HB=\dfrac{72}{5}\Rightarrow BC=\dfrac{128+72}{5}=40\left(cm\right)\)
\(\Rightarrow AC=\sqrt{BC ^2-AB^2}=\sqrt{40^2-24^2}=32\)
Ta có: \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{24.32}{40}=\dfrac{96}{5}\left(cm\right)\)
\(\dfrac{HB}{HC}=\dfrac{9}{16}\Rightarrow HC=\dfrac{16}{9}HB\)
Áp dụng hệ thức lượng:
\(AB^2=HB.BC=HB\left(HB+HC\right)\)
\(\Leftrightarrow24^2=HB.\left(HB+\dfrac{16}{9}HB\right)\)
\(\Rightarrow HB^2=\dfrac{5184}{25}\Rightarrow HB=\dfrac{72}{5}\left(cm\right)\)
\(HC=\dfrac{16}{9}HB=\dfrac{128}{5}\) (cm)
\(BC=HB+HC=40\) (cm)
\(AC=\sqrt{BC^2-AB^2}=32\) (cm)
\(AH=\dfrac{AB.AC}{BC}=\dfrac{96}{5}\left(cm\right)\)